# Intuition about the values of Cauchy integrals on the unit circle

Considering the Cauchy integral $$\oint \frac{f(x)}{(x-z)} dx$$ over the unit circle. The integral equals zero if $z$ is outside the circle, yet it has a value if $z$ is inside the circle.

Why is this the case if one just looks at it computationally (without using the Fundamental Theorem of Calculus, etc.). I.e., what can be said along the lines of: when $z$ in in the interior, this and this happens to give a value; whereas when it's exterior, that and that happens to result in $0$. Thanks!

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Well what you've said is not true for all functions $f$. Perhaps you had a specific one in mind, or for this question it suffices to consider $f(x)=1$ ? – Ragib Zaman Aug 28 '11 at 14:41
The assumption "$f$ holomorphic" (in a region that includes the unit circle) should be added. – leonbloy Aug 28 '11 at 14:53
+1 for the way the question aims at building intuition. – Ben Blum-Smith Aug 28 '11 at 15:31
If you look at the integral "computationally" something terrible is happening: At the moment when $z$ crosses the unit circle the integral becomes undefined; it then looks like a real integral $\int_{-1}^1{1\over x}\ dx$. This implies that there is no connection between the values of the integral for $z$ inside and outside the unit circle. – Christian Blatter Aug 28 '11 at 16:08

Here is a different (much more imprecise and intuitive, but hopefully illuminating, and I believe along the lines of what you were asking) angle on it.

The theorems of complex analysis are so powerful (for computing integrals in terms of residues etc.) that I sometimes forget integrals are really limits of sums. Thinking of the complex integral in these terms:

As in @Did's answer, take $f=1$; everything you need is contained in this example. To get a general feel for what happens when $z$ is inside the circle, take the case that $z=0$. Then the integral is $\int_C \frac{1}{x}dx$ where $C$ is the unit circle, oriented counterclockwise. Use the standard calculus fudge of imagining this as the sum of infinitely many infinitesimal complex numbers. Each infinitesimal summand is $dx/x$ where $x$ lies on the unit circle. $dx$ is a little bit of circle, i.e. an infinitesimal vector tangent to the circle. $x$ is the radius of the circle that ends at this tangent. They are at right angles, so the ratio $dx/x$ is a multiple of $i$. Since the circle is oriented ccw, it is a positive multiple of $i$ and because $x$ has unit length, it has the magnitude of $dx$. So each element $dx/x$ is really an arc length element times $i$. Adding all these up we get the full arc length times $i$, i.e. $2\pi i$.

If $z$ is not zero but still inside the circle, the calculation is less nice, but an essentially similar thing happens: $z$ is always to the left of you as you walk ccw along the circle, so the angle between $dx$ and $x-z$ will stay in a confined range (less than 180 degrees). Then the sum of the elements $dx/(x-z)$ is nonzero since the arguments being close together prevents them from cancelling each other. (In fact it is always $2\pi i$ for this function, but the point is that it is not zero.)

If $z$ is outside the circle, the key difference is that now as you travel along the circle you can't keep $z$ on your left the whole time, so the elements $dx/(x-z)$ will be forced to point in all directions. For example take $z=2$. Then the integral is $\int_C \frac{1}{x-2}dx$. Imagining $x$ somewhere on the circle now, $dx$ is again a little bit of circle, and now $x-2$ is the vector from $2$ to $x$. $\frac{dx}{x-2}$ has a magnitude based on the (changing) magnitude of $x-2$ and its argument is given by the angle between $dx$ and $x-2$. As $x$ goes around the circle, you can see that $dx/(x-2)$ ends up pointing in all directions. This is how the sum of all of them can cancel out. The details of how the cancellation takes place aren't what I'm stressing - you can write out $x-2$ and $dx$ in terms of real and imaginary parts based on a parametrization like $x=\cos\theta + i\sin\theta$ and crunch out the integral to get zero, but I didn't find that very illuminating, and couldn't see a simpler geometric argument that they cancel out perfectly. The main point is just that because the circle does not enclose $2$, as $x$ goes along the circle, $2$ is not strictly to your left as you walk around the circle, so the value of the elements $dx/(x-2)$ have to double back on themselves.

A good deal of this discussion has been specific to the function $f=1$, but the main point is actually extremely general: If $z$ inside the loop, then going around the loop you can keep $z$ on your left the whole time, and an integral of $f/(x-z)$ can exploit this to accumulate some nonzero value. If $z$ is outside the loop, then as you go around the loop, $z$ will be on your left sometimes and your right sometimes, and under the right geometric conditions on $f$ (specifically, holomorphic in a region containing the unit disc), this forces all the elements $fdx/(x-z)$ to cancel.

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The case $f=1$ is sufficient to understand what is going on, so let us assume this, and let us try to expand the function $u(x)=1/(x-z)$ as a series. If $|z|<1$, $|z/x|<1$ hence $$u(x)=\frac1x\sum\limits_{n=0}^{+\infty}\left(\frac{z}x\right)^n.$$ Since this series converges normally on $|x|=1$, it is legal to integrate $u(x)$ term by term. One gets $0$ for every term except $n=0$, for which one gets $2\text{i}\pi$.

Now, if $|z|>1$, $|z/x|>1$ hence the previous expansion is not valid. Fortunately, $|x/z|<1$ hence $$u(x)=-\frac1z\sum\limits_{n=0}^{+\infty}\left(\frac{x}z\right)^n.$$ Here again, the convergence of the series is normal on $|x|=1$ hence it is legal to integrate $u(x)$ term by term. One gets $0$ for every term with no exception.

In the end, the idea is that the integral $\displaystyle I_n=\oint\frac{\text{d}x}{x^n}$ is $I_n=0$ for every integer $n$ except $n=1$ for which $I_1=2\text{i}\pi$ and that the integral one considers in the question involves $I_{1}$ when $|z|<1$ but not when $|z|>1$.

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Dear Did - When I first asked this question two years ago, I had just begun self-studying with no formal math education. Now I am studying CA again and I can appreciate this most beautiful answer. Thanks and regards, – TheBirdistheWord Jul 14 '13 at 12:00
@Andrew Thanks. $\langle$ Blushes $\rangle$. – Did Jul 14 '13 at 12:11

This phenomenon merits attention, indeed. On one hand, we certainly want to assimilate Cauchy's formula into our intuition. On the other hand, it is an example of strange phenomena in integral representations of (holomorphic?) functions.

That is, the "Cauchy kernel" $K(z,w)=\frac{1}{z-w}$ has the property asserted by Cauchy's formula, namely, that integrating against $f(z)$ on a circle containing $w$ produces essentially $f(w)$, while for $w$ outside the circle, we get $0$. Examination of the integral's behavior as $w$ crosses the circle is not so elementary, and is usually and reasonably not discussed in introductory situations. Indeed, even a radial limit is at best a principal value, etc. And there is a definite discontinuity as $w$ crosses the circle. I think it is fair to say that the disparity of behaviors inside and outside the circle is not obvious from the appearance of the integral.

To put this behavior in a larger context, one might consider other kernels $K(z,w)$ and integrate over circles or lines. Obviously many cases reduce to Cauchy's, or, the same thing, to "residues".

There are examples with less discontinuity than in Cauchy's basic example, although maybe it's not trivial to produce elementary ones. That there are non-trivial possibilities was forcefully brought to my attention in some basic analytic number theory/automorphic forms business (perhaps a fancier issue than directly desired by the questioner, but there is a further, potentially useful point): part of a resolvent or fundamental solution in automorphic forms includes an integral $$\int_{\Re(z)=1/2} \frac{E_{1-z}\,E_z\,dz}{(z-w)(z-1+w)}$$ where $E_z$ is an Eisenstein series (with its own issues/properties...). This expression converges nicely for $\Re(w)>1/2$, and (here comes the deceitful part) the integrand is symmetric under $w\rightarrow 1-w$ and in fact is continuous as $\Re(w)\rightarrow 1/2^+$. So, yes, that integral extends to a meromorphic function on the whole plane. However, as $w$ passes through the line $\Re(w)=1/2$, its nature changes in several subtle, technical ways, among which is the violently non-intuitive point that the meromorphic continuation is not given by that integral! (What?!) Extra terms are picked up!!! The argument is by residues, that is, by Cauchy's formula...

The details are not so much my point, but that the thought-provoking aspects of Cauchy's formula have considerable not-so-innocent potential!

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@Didier Piau... Thanks for copy-edits! [I did try to ascertain your for-yourself email, but found some obstacles... therefore I say "thanks" in front of the world... :) ...] – paul garrett Aug 30 '11 at 2:04
you are welcome. – Did Aug 30 '11 at 8:30
Dear Prof. Garrett. Thanks you for considering my question and adding substantial additional insight. I have been remiss in formally expressing my appreciation, which I always deeply felt, for anyone who even considers my (and really all) questions here. Thanks and best regards, – TheBirdistheWord Sep 15 '11 at 21:49